How to integrate $\int_0^\infty e^{-ty^2} \sin t dt$

Question

My book suggests that I do some sort of limiting

$\lim_{A \to \infty} \int_0^A e^{-ty^2} \sin t d t$

But I'm not getting anywhere.

Answer 1

An alternative way is to recognize that

$$\int_0^{\infty} dt \, e^{-b t} = \frac{1}{b}$$

when $\Re{b} \gt 0$. Then write

$$\int_0^{\infty} dt \, e^{-t y^2} \, \sin{t} = \Im{\left [ \int_0^{\infty} dt \, e^{-(y^2 - i) t} \right]} = \Im{\left[\frac{1}{y^2-i}\right]} = \frac{1}{y^4+1}$$

Answer 2

Starting with the initial integral. You can integrate it by parts. $$\int_0^A e^{-ty^2} \sin t d t=-\int_0^A e^{-ty^2} d \cos(t)=-\left(e^{-ty^2}\cos(t)\bigg|_0^A -(-y^2)\int_0^A \cos(t)e^{-ty^2} dt \right)$$ $$\int_0^A e^{-ty^2} \sin t d t=-e^{-ty^2}\cos(t)\bigg|_0^A -y^2\int_0^A e^{-ty^2} d\sin(t)=-e^{-ty^2}\cos(t)\bigg|_0^A -y^2\left(e^{-ty^2}\sin(t)\bigg|_0^A -(-y^2)\int_0^A \sin(t)e^{-ty^2} dt \right) $$ $$\int_0^A e^{-ty^2} \sin t dt=1-e^{-Ay^2}(\cos(A)+y^2\sin(A))-y^4\int_0^A e^{-ty^2} \sin t dt$$ $$(1+y^4)\int_0^A e^{-ty^2} \sin t dt=1-e^{-Ay^2}(\cos(A)+y^2\sin(A))$$ $$\int_0^A e^{-ty^2} \sin t dt=\frac{1-e^{-Ay^2}(\cos(A)+y^2\sin(A))}{(1+y^4)}$$ And finally takong the limit one can obtain the answer: $$\lim_{A \to \infty} \int_0^A e^{-ty^2} \sin t d t=\frac{\lim_{A \to \infty} (1-e^{-Ay^2}(\cos(A)+y^2\sin(A))) }{1+y^4}=\frac{1}{1+y^4}$$

Answer 3

you can use Euler formula $e^{i\pi}=\cos(x)+i\sin(x)$

Answer 4

Well, I guess there some other interesting ways to solve the problem. One more try. You can look at that integral as the Laplace transform: $$F(s) =\int_0^{\infty} e^{-st} f(t) \,dt$$ obtained for $s=y^2$.
So you have $f(t)=\sin(t), \ s=y^2$. It is well known that the Laplace transform of $ \sin(\omega t)$ is $ { \omega \over s^2 + \omega^2 } $ . So just set $\omega=1$ and one will obtain: $$\int_0^\infty e^{-ty^2} \sin t dt=\frac{1}{y^4+1}$$